Do Hamiltonians Fully Represent All Observables in Quantum Mechanics?

[kye]

It is said that each observable like position or momentum is represented by a Hermitian operator acting on the state space. And the Hamiltonian is the total energy of the system, kinetic and potential.. so it means the Hamiltonians encode or encompass the energy of all observables (like position, spin, charges, momentum, etc.) at the same instance? Or do you analyze each Hamiltonian separately for each observable and add them together?

Second inquiry. In Wikipedia it is stated that "As it turns out, analytic solutions of the Schrödinger equation are available for only a very small number of relatively simple model Hamiltonians, of which the quantum harmonic oscillator, the particle in a box, the hydrogen molecular ion, and the hydrogen atom are the most important representatives. Even the helium atom - which contains just one more electron than does the hydrogen atom - has defied all attempts at a fully analytic treatment.". Does it mean we can only get the complete Hamiltonian (combination of all observables) for only hydrogen atom and barely for helium? Meaning there is no way to get the Hamiltonian of molecules?

 

[atyy]

It is said that each observable like position or momentum is represented by a Hermitian operator acting on the state space. And the Hamiltonian is the total energy of the system, kinetic and potential.. so it means the Hamiltonians encode or encompass the energy of all observables (like position, spin, charges, momentum, etc.) at the same instance? Or do you analyze each Hamiltonian separately for each observable and add them together?

The Hamiltonian is a property of the system, which determines time evolution of its state (the wave function or state vector).

Observables correspond to Hermitian operators. Since the Hamiltonian is also a Hermitian operator, it corresponds to an observable - the energy.

There is no such thing as the Hamiltonian for an observable.

Second inquiry. In Wikipedia it is stated that "As it turns out, analytic solutions of the Schrödinger equation are available for only a very small number of relatively simple model Hamiltonians, of which the quantum harmonic oscillator, the particle in a box, the hydrogen molecular ion, and the hydrogen atom are the most important representatives. Even the helium atom - which contains just one more electron than does the hydrogen atom - has defied all attempts at a fully analytic treatment.". Does it mean we can only get the complete Hamiltonian (combination of all observables) for only hydrogen atom and barely for helium? Meaning there is no way to get the Hamiltonian of molecules?

We have the complete Hamiltonian for all these systems. However, we do not have exact solutions, for example, for their ground state wave functions (ie. the state of lowest energy). However, we can get pretty good approximations for Helium. People still research how to get the wave functions for complicated systems whose Hamiltonian is known.

 

[kye]

The Hamiltonian is a property of the system, which determines time evolution of its state (the wave function or state vector).

Observables correspond to Hermitian operators. Since the Hamiltonian is also a Hermitian operator, it corresponds to an observable - the energy.

There is no such thing as the Hamiltonian for an observable.



We have the complete Hamiltonian for all these systems. However, we do not have exact solutions, for example, for their ground state wave functions (ie. the state of lowest energy). However, we can get pretty good approximations for Helium. People still research how to get the wave functions for complicated systems whose Hamiltonian is known.

If Hamiltonian is a Hermitian operator that corresponds to the energy observable. What is the corresponding Hermitian for other observables? It's not called Hamiltonian anymore?

Also how many observables are possible and what are they?

 

[atyy]

If Hamiltonian is a Hermitian operator that corresponds to the energy observable. What is the corresponding Hermitian for other observables? It's not called Hamiltonian anymore?

For other observables, the Hermitian operator is usually also called by the same name as the observable. For example, the position observable corresponds to the position operator. Energy is the only observable I know of whose operator (the Hamiltonian) has a different name. It is certainly not wrong to call the Hamiltonian the energy operator.

Also how many observables are possible and what are they?

I don't know the answer to the first part, but I think that there are usually more mathematically possible observables than can be practically measured. The observables depend on the system. In non-relativistic quantum mechanics of single particles whose state is specified by a wave function ψ(x), some observables are position, momentum, energy. If the system is an electron, we could also ask about other observables such as the z-direction of its intrinsic spin, or the x-direction of its intrinsic spin.

 

[kye]

We have the complete Hamiltonian for all these systems. However, we do not have exact solutions, for example, for their ground state wave functions (ie. the state of lowest energy). However, we can get pretty good approximations for Helium. People still research how to get the wave functions for complicated systems whose Hamiltonian is known.

Why don't we have exact solutions even if we have complete Hamiltonian for all these systems? Do you mean complete as in we know the energy but not the exact energy, then it's not complete, so how could you say it's complete. Or is it like Perturbation in QED where there is no exact solution due to the virtual particles in the ground state of the hydrogen? or is it because we don't have computers fast enough to solve for it?

For other observables, the Hermitian operator is usually also called by the same name as the observable. For example, the position observable corresponds to the position operator. Energy is the only observable I know of whose operator (the Hamiltonian) has a different name. It is certainly not wrong to call the Hamiltonian the energy operator.

Does the Hamiltonian take into account all the momentums from orbitals, spins, etc.? and is there analyzer or spectroscopy that can detect the Hamiltonian energy? And if it can, can it tell what percentage came from the spins or orbitals or others (the quantum numbers)? If not.. how do you tell the contributaitons from spin or orbital?

I don't know the answer to the first part, but I think that there are usually more mathematically possible observables than can be practically measured. The observables depend on the system. In non-relativistic quantum mechanics of single particles whose state is specified by a wave function ψ(x), some observables are position, momentum, energy. If the system is an electron, we could also ask about other observables such as the z-direction of its intrinsic spin, or the x-direction of its intrinsic spin.

Don't other observables also contain either kinetic or potential energy that the Hamiltonian have or do other observables are filtered such that they only treat the non-Hamiltonian parts?

 

[atyy]

Why don't we have exact solutions even if we have complete Hamiltonian for all these systems? Do you mean complete as in we know the energy but not the exact energy, then it's not complete, so how could you say it's complete. Or is it like Perturbation in QED where there is no exact solution due to the virtual particles in the ground state of the hydrogen? or is it because we don't have computers fast enough to solve for it?

By having no "analytic solution", it is meant that we need a computer to determine, for example, the stationary wave function that has the lowest energy. If a computer solves for it, it is an approximate solution. In many cases, computers are not fast enough to provide us with good approximate solutions.

Does the Hamiltonian take into account all the momentums from orbitals, spins, etc.? and is there analyzer or spectroscopy that can detect the Hamiltonian energy? And if it can, can it tell what percentage came from the spins or orbitals or others (the quantum numbers)? If not.. how do you tell the contributaitons from spin or orbital?

The Hamiltonian (and the state space) are the complete specification of the system, so yes it would take into account all interactions between spins etc.

There is no general procedure for figuring out the Hamiltonian of a system. Usually we have experiments from which guesses or hypotheses about the Hamiltonian are made. We test the hypothesis by doing more experiments and modifying our hypothesis if the predictions made by the hypothesis are wrong.

Don't other observables also contain either kinetic or potential energy that the Hamiltonian have or do other observables are filtered such that they only treat the non-Hamiltonian parts?

Yes, other observables can be a part of the Hamiltonian.

 

[kye]

By having no "analytic solution", it is meant that we need a computer to determine, for example, the stationary wave function that has the lowest energy. If a computer solves for it, it is an approximate solution. In many cases, computers are not fast enough to provide us with good approximate solutions.




The Hamiltonian (and the state space) are the complete specification of the system, so yes it would take into account all interactions between spins etc.

There is no general procedure for figuring out the Hamiltonian of a system. Usually we have experiments from which guesses or hypotheses about the Hamiltonian are made. We test the hypothesis by doing more experiments and modifying our hypothesis if the predictions made by the hypothesis are wrong.



Yes, other observables can be a part of the Hamiltonian.

Thanks. In Quantum field theory, the points are not field values. Not only that. It's not even observables or state vector... but another operator acting on the state vector (just to get your context.. what is your understanding of "state vector" as it relates to "observables"?). And what particular operator is acting in QFT that create and annihilate particles. Maybe a separate creation and annihilation operator which is extra to QM? So every point of the field has this same C & A operator or can you also use hamiltonian or spin, momentum operator acting on the points of state vector in QFT?

 

[atyy]

Thanks. In Quantum field theory, the points are not field values. Not only that. It's not even observables or state vector... but another operator acting on the state vector (just to get your context.. what is your understanding of "state vector" as it relates to "observables"?). And what particular operator is acting in QFT that create and annihilate particles. Maybe a separate creation and annihilation operator which is extra to QM? So every point of the field has this same C & A operator or can you also use hamiltonian or spin, momentum operator acting on the points of state vector in QFT?

Let's keep to non-relativistic QFT (there's no conceptual obstacle to generalizing to relativistic QFT). It is just a way of doing quantum mechanics (with the non-relativistic Schroedinger equation) with many identical particles. QFT is exactly QM, just from a different viewpoint. The method of translating between the viewpoints is called "second quantization".

So I would suggest you learn QM well first.

There is an analogue of the creation and destruction operators in QFT in the harmonic oscillator in QM.

http://www.anst.uu.se/thokl773/kf2011/kf2011-lecture03-osci.pdf
http://www.oberlin.edu/physics/dstyer/QM/Assignments/LadderOperators.pdf

 

[kye]

By having no "analytic solution", it is meant that we need a computer to determine, for example, the stationary wave function that has the lowest energy. If a computer solves for it, it is an approximate solution. In many cases, computers are not fast enough to provide us with good approximate solutions.




The Hamiltonian (and the state space) are the complete specification of the system, so yes it would take into account all interactions between spins etc.

There is no general procedure for figuring out the Hamiltonian of a system. Usually we have experiments from which guesses or hypotheses about the Hamiltonian are made. We test the hypothesis by doing more experiments and modifying our hypothesis if the predictions made by the hypothesis are wrong.



Yes, other observables can be a part of the Hamiltonian.

atty, what quantum operators or principles that is related to thermodynamics in chemical system? I know that in temperature, it is from the ensemble of atoms and molecules that have Brownian motion... but the atoms and molecules are not classical balls, but are quantum states.. so how do you treat quantum states that bounced off one another? the point is the behavior may not be like billiard boards bounching off one another.

 

[atyy]

atty, what quantum operators or principles that is related to thermodynamics in chemical system? I know that in temperature, it is from the ensemble of atoms and molecules that have Brownian motion... but the atoms and molecules are not classical balls, but are quantum states.. so how do you treat quantum states that bounced off one another? the point is the behavior may not be like billiard boards bounching off one another.

I'm not so familiar with chemistry, but the basic idea is the same as classical thermodynamics.

Classical thermodynamics deals with things that are "coarse grained" variables like pressure and temperature, whereas the particles in a gas have microscopic variables like position and momentum. The relationship between microscopic and coarse grained variable is given by classical statistical mechanics. Quantum statistical mechanics is essentially the same as the classical case, only that the notion of state is different - in the classical case the state is specified by position and momentum, in the quantum case the state is a vector in Hilbert space.

 

[kye]

I'm not so familiar with chemistry, but the basic idea is the same as classical thermodynamics.

Classical thermodynamics deals with things that are "coarse grained" variables like pressure and temperature, whereas the particles in a gas have microscopic variables like position and momentum. The relationship between microscopic and coarse grained variable is given by classical statistical mechanics. Quantum statistical mechanics is essentially the same as the classical case, only that the notion of state is different - in the classical case the state is specified by position and momentum, in the quantum case the state is a vector in Hilbert space.

maybe temperature is not used for electrons because it would take something hotter than the center of the sun to make the electrons change to higher orbitals whereas it would take just a gamma ray to do that because of more focused energy. is this what you have in mind?

how do you view molecules where the atoms are in superposition? Even if you heat an object, the atoms would have the energy subdivided or partitioned such that the electrons won't be affected (coarsed grained), is there no way for temperature to change the quantum state of atoms? or what are the known ways I'm not aware of?

 

[atyy]

Temperature is used for electrons. Both classically and quantum mechanically the relationship between temperature and microscopic variables are given by statistical mechanics.

 

[kye]

I thought the microscopic variables are only for atoms.. in a box.. to say produce temperature.. I read that you can dislodge an electron from an orbital either by strong gamma ray or strong heat like in Big Bang where the electrons are ionized.. so what situations do you use temperature for electrons when they are tightly bound to the atom?

 

[atyy]

In the BCS theory of low-temperature superconductivity, the main degrees of freedom are interacting electrons and phonons. This interacting system of electrons and phonons is usually treated using quantum statistical mechanics in the grand canonical ensemble. The grand canonical ensemble assigns the system a temperature and is able to predict that the system loses superconductivity above a certain temperature.

 

[kye]

In the BCS theory of low-temperature superconductivity, the main degrees of freedom are interacting electrons and phonons. This interacting system of electrons and phonons is usually treated using quantum statistical mechanics in the grand canonical ensemble. The grand canonical ensemble assigns the system a temperature and is able to predict that the system loses superconductivity above a certain temperature.

What else beside the BCS theory of low-temp superconductivity treat the electrons using statistic mechanics?

 

[kye]

Let's keep to non-relativistic QFT (there's no conceptual obstacle to generalizing to relativistic QFT). It is just a way of doing quantum mechanics (with the non-relativistic Schroedinger equation) with many identical particles. QFT is exactly QM, just from a different viewpoint. The method of translating between the viewpoints is called "second quantization".[/url]

QFT is operator acting state vector on a "point" of space, whereas QM is just state vector. To analyze the Hamiltonian of say the electron and using QFT (just for sake of discussion), how do you remove the operator on the state vector in QFT.. maybe by making the operator zero so it becomes just state vector?